poj 3140

最新推荐文章于 2019-03-20 00:22:55 发布

weixin_30767921

最新推荐文章于 2019-03-20 00:22:55 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zhaozhe/archive/2011/05/10/2042075.html

简单题

代码:

#include<iostream>
#include<fstream>
#include<vector>

using namespace std;

vector<long long> edge[100001];
long long value[100001];
int v[100001];
long long dp[100001];
long long n,m;

long long ok(long long s){
	if(s<0) return -s;
	return s;
}

long long solve(long long s){
	long long i;
	v[s]++;
	dp[s]=value[s];
	for(i=0;i<edge[s].size();i++)
	{
		if(v[edge[s][i]]==0)
			dp[s]+=solve(edge[s][i]);
	}
	return dp[s];
}

void read(){
//	ifstream cin("in.txt");
	long long i,j,k=0,s,t;
	while(cin>>n>>m)
	{
		k++;if(n==0&&m==0) return ;
		for(i=1;i<=n;i++)
			edge[i].clear();
		long long sum=0;
		
		memset(v,0,sizeof(v));
		for(i=1;i<=n;i++)
		{
			cin>>value[i];
			sum+=value[i];
		}
		for(i=1;i<=m;i++)
		{
			cin>>s>>t;
			edge[s].push_back(t);
			edge[t].push_back(s);
		}
		solve(1);
		long long ans=ok(sum-2*dp[1]);
		for(i=2;i<=n;i++)
			if(ans>ok(sum-2*dp[i]))
				ans=ok(sum-2*dp[i]);
		printf("Case %lld: %lld\n",k,ans);
	}
	


}

int main(){
	read();
	return 0;
}

转载于:https://www.cnblogs.com/zhaozhe/archive/2011/05/10/2042075.html